# Understanding *x ,= lst

I'm going through some old code trying to understand what it does, and I came across this odd statement:

```
*x ,= p
```

`p`

is a list in this context. I've been trying to figure out what this statement does. As far as I can tell, it just sets `x`

to the value of `p`

. For example:

```
p = [1,2]
*x ,= p
print(x)
```

Just gives

```
[1, 2]
```

So is this any different than `x = p`

? Any idea what this syntax is doing?

`*x ,= p`

is basically an obfuscated version of `x = list(p)`

using extended iterable unpacking. The comma after `x`

is required to make the assignment target a tuple (it could also be a list though).

`*x, = p`

**is** different from `x = p`

because the former creates a *copy* of `p`

(i.e. a new list) while the latter creates a *reference* to the original list. To illustrate:

```
>>> p = [1, 2]
>>> *x, = p
>>> x == p
True
>>> x is p
False
>>> x = p
>>> x == p
True
>>> x is p
True
```

From: stackoverflow.com/q/43190992

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